Determination of differential pencils with spectral parameter dependent boundary conditions from interior spectral data

Second‐order differential pencils L(p,q,h0,h1,H0,H1) on a finite interval with spectral parameter dependent boundary conditions are considered. We prove the following: (i) a set of values of eigenfunctions at the mid‐point of the interval [0,π] and one full spectrum suffice to determine differential pencils L(p,q,h0,h1,H0,H1); and (ii) some information on eigenfunctions at some an internal point b∈(π2,π) and parts of two spectra suffice to determine differential pencils L(p,q,h0,h1,H0,H1). Copyright © 2013 The Authors. Mathematical Methods in the Applied Sciences published by John Wiley & Sons, Ltd.

Inverse problem for interior spectral data of the differential operator lies in reconstructing this operator by some eigenvalues and information on eigenfunctions at some an internal point in the interval considered. The similar problems for the Sturm-Liouville operators [25,26] and differential pencils [27] were studied.
As far as I know, the inverse problem of interior spectral data for differential pencils L with spectral parameter dependent boundary conditions has not been considered before. The aim of this paper is to give two uniqueness theorems from some eigenvalues and information on eigenfunctions at some an internal point in the interval OE0, provided that the spectrum is simple. The results obtained are new and a generalization of the well-known one for the classical Sturm-Liouville operator, which was studied in [25], for a special case that p.x/ Á 0, h 1 D H 1 D 0, and the parameters h and H are fixed. The results in this paper are also a generalization of theorems because of Yang and Guo [27], where authors consider a special case that h 1 D H 1 D 0 and assume either p.x/ or q.x/ is a prior known.
In fact, the last restriction is unnecessary. The proof presented here follows Mochizuki and Trooshin's proof in outline, but the results obtained are more general.

Main results
The eigenvalue asymptotics for pencils L are studied in [15], and it is shown that eigenvalues are not necessarily real, and all but a finite number of the eigenvalues are algebraically simple. It is well-known that the eigenvalues of pencils L consist of the sequence n , n 2 A def D f˙0,˙1,˙2, g, and large eigenvalue n satisfies the classical asymptotic form To simplify the calculations later, we will assume that the eigenvalues n are all simple. Denote by n and y n .x/, n 2 A, eigenvalues and corresponding eigenfunctions of the differential pencils L, respectively. We introduce the quantities(derivatives of the logarithm of jy n .x/j) Consider a second differential pencils 8 where e h k , e H k , e q.x/ and e p.x/ have the same properties of h k , H k , q.x/ and p.x/, k D 0, 1. We agree that if a certain symbol ı denotes an object related to L.p, q, h 0 , h 1 , H 0 , H 1 /, then e ı will denote an analogous object related to L e p,e q, e h 0 , e h 1 , e H 0 , e H 1 .
Let l.n/, r.n/ be sequences of natural numbers with properties l.n/ D n 1 . The solution of inverse problem in Theorem 2.1 is not unique without condition Ä n 2 D e Ä n 2 , because single spectrum can not determine the pencils L. In particular, when y n . 2 / D 0, equation Ä n 2 D e Ä n 2 is replaced by y n . 2 /e y 0 n . 2 / D y 0 n . 2 /e y n . 2 /. Theorem 2.3 Let l.n/, r.n/ and b 2 . 2 , / be such that 1 > 2b 1, 2 > 2 2b . If for any n 2 A,

Integral representation of products
In this section, we shall study two products of the eigenfunctions of two differential pencils, see (3.6). The result will be used to derive an integral equation for the functions p.x/ e p.x/ and q.x/ e q.x/, which does not involve the parameter .
Let y 1 .x, / be the solution of equation l y 1 .x, / D 0 satisfying the initial conditions y 1 .0/ D 1 and y 0 1 .0/ D h 0 , then where˛.x/ D R x 0 p.t/dt, and kernels A 1 .x, t/ and B 1 .x, t/ are solutions of the following problem [19]: 8 < : For each 2 C, ¤ 0, let y 2 .x, / be the solution of equation l y 2 .x, / D 0 satisfying the initial conditions y 2 .0/ D 0 and y 0 2 .0/ D 1, then where the kernels A 2 .x, t/, B 2 .x, t/ are the solution of the problem 8 < : Let y.x, / be the solution to equation l y.x, / D 0 with the initial conditions y.0/ D 1 and y 0 .0/ D .i h 1 C h 0 /, then from (3.1) and (3.2), we have Moreover, we obtain Simple calculations show that the characteristic equation of the pencils L can be reduced to !. / D 0, where Denote G ı D f : j k !j ı, k D 0,˙1,˙2, g, ı > 0. Using the known method (see, e.g. [4]), one can prove the following estimates for sufficiently large j j: Similarly, for the solution e y.x, / of equation e l e y.x, / D 0 with the initial conditions e y.0/ D 1 ande y 0 .0/ D i e h 1 C e h 0 , there has the following analogous result: Next, using (3.3) and (3.5), and by extending the range of A.x, t/, e A.x, t/ evenly with respect to the argument t and B.x, t/, e B.x, t/ oddly with respect to the argument t and some straightforward calculations, for brevity denoting

Completion of proofs
In this section, we shall complete the proofs of Theorems 2.1 and 2.3. The basic idea is to translate the integral equation (4.1) into an inhomogeneous integral equations that are independent of and then show, step by step, that the inhomogeneous terms must vanish. Now, we can give the proofs of theorems in this work. Using the initial conditions at 0, then it yields OE.e q q/ C 2 .e p p/ ye ydx D 0.
By use of the Riemann-Lebesgue Lemma, we see that the limit of the left-hand side of the earlier equality exists as ! 1, 2 R. Thus, we obtain that the constant C D 0. So, we have proved  Moreover, from H. / D 0 for all complex number and by use of the Riemann-Lebesgue Lemma as ! 1, 2 R, we obtain that Thus, we have By changing the order of integration, (4.9) can be rewritten as where By integration by parts in (4.11), we have (4.12) Moreover, from H. / D 0 for all complex number and by use of the Riemann-Lebesgue Lemma as ! 1, 2 R, we obtain that and From the definitions of R 1 .t/, R 2 .t/, T 1 .t/, T 2 .t/ by (4.11), we can infer Substituting (4.10) into (4.16), together with P. =2/ D 0, it follows that 8 < : and But this is a homogeneous Volterra integral equation, and its solution is identically zero a.e. Thus, we have obtained and H k D e H k .k D 0, 1/, we should repeat the earlier argument for the supplementary problem 8 < :  Using the same method in [27] (pp. 291-292), we can show that G. / D 0 on the whole complex plane. As we already mentioned, if G. / Á 0, then the conclusion of Lemma is true.
(2) Note that the interval OEb, can be converted to an interval OE0, b by a transformation of variable x 7 ! x. To prove (2), we should repeat arguments in part (1)